Optimum MSE for neural networks

John Williams answered . 2024-12-20 18:10:20

 [ I N ]    = size(x)
 [ O N ]    = size(t) = size(y)
 Ntrn      = No. of training examples ( Ntrn ~ 0.7*N is default )
 Ntrneq    = Ntrn*O  % No. of training equations
 MSEtrn00  = mean(var(ttrn',1))% avg training target variance
 SSEtrn    = sse(ttrn-ytrn)
 MSEtrn    = SSEtrn/Ntrneq
 NMSEtrn   = MSEtrn/MSEtrn00
 Rsqtrn    = 1 - NMSEtrn

Adjustments ("a") for degrees of freedom lost when evaluating the net with the same data that was used to estimate the weights:

 H         = number of hidden nodes
 Nw       = (I+1)*H+(H+1)*O  %  No. of unknown weights
 Ndof     =  Ntrneq - Nw     % No. of DOF
 MSEtrn00a = mean(var(ttrn',0))

 MSEtrna   = SSEtrn/Ndof
 NMSEtrna  = MSEtrna/MSEtrn00a
 Rsqtrna   = 1 - NMSEtrna

For many problems an appropriate training goal is

 R2sqtrna >= 0.99

or

 MSEtrn <= MSEtrngoal = 0.01*max(Ndof,0)*MSEtrn00a/Ntrneq

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